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3 years ago

Simplify each expression. (8 - 6)^5 times (2) ^-4+ (2)^0

Mathematics

2 answers:

natta225 [31]3 years ago

7 0

Answer: 1. 32. 2.13

Step-by-step explanation:

hodyreva [135]3 years ago

3 0

Answer:

Step-by-step explanation:

$(8-6)^5 * (2^{-4}+(2)^0)\\=2^5 *(\frac{1}{16} +1)\\=32*(\frac{17}{16})\\ =2*17\\=34$

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HELP ASAP 10 POINTS TO BRAINLIEST

Brilliant_brown [7]

Answer:

15%

Step-by-step explanation:

The cost of 2 adults and 2 children without the discount is

2 adults = 2 (55) = 110

2 children = 2 (45) = 90

Total cost = 110+90 = 200

The original price is 200 and the new price is 170

Percentage discount = (Original price - new price)/ original price * 100%

= (200-170)/200 * 100%

= 30/200 * 100%

= .15 * 100%

= 15%

6 0

3 years ago

In science class, Priscilla heats water and measures the temperature of the water every 2 minutes\or data are shown in the table

Dima020 [189]

Answer:

37.5 °C

Step-by-step explanation:

The temperature appears to be increasing linearly at the rate of 5 °C in 2 minutes, or 2.5 °C per minute.

1 minute after 4 minutes, when the temperature is 35 °C, we expect it to be 2.5 °C higher, or 37.5 °C.

8 0

2 years ago

In a shipment of 22 smartphones, 2 are defective. How many ways can a quality control inspector randomly test 4 smartphones, of

Eddi Din [679]

Answer:

Therefore the required ways are =190

Step-by-step explanation:

Combination: Combination is the number of selection of items from a collection of items where the order of selection does not matter.

Total number of smartphones =22

Defective phone = 2

Non defective phone = (22-2) =20

The control inspector randomly test 4 smartphones of which 2 are defective.

Non defective phone = 2

The ways to select 2 non defective phone is

$^{20}C_2=\frac{20!}{2!(20-2)!} =\frac{20!}{2!18!}=\frac{19\times 20}{2} =190$

The ways to select 2 defective phone is= $^2C_2=1$

Therefore the required ways are = (190×1) =190

5 0

3 years ago

What is the length of line segment KJ?

Karolina [17]

Please consider the attached file.

We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.

We will use Pythagoras theorem to solve for KJ as:

$KJ^2=KM^2+MJ^2$

$KJ^2=6^2+3^2$

$KJ^2=36+9$

$KJ^2=45$

Now we will take positive square root on both sides:

$\sqrt{KJ^2}=\sqrt{45}$

$KJ=\sqrt{9\cdot 5}$

$KJ=3\sqrt{5}$

Therefore, the length of line segment KJ is $3\sqrt{5}$ and option D is the correct choice.

5 0

3 years ago

WVXZ is a parallelogram. Given that <ZXV = 70° and <WZX =4y+34, find y and m<Z

rjkz [21]

Answer:

WZX = 110

y = 19

Step-by-step explanation:

We can create the following equation to solve for y.

$4y + 34 + 70 = 180$

Reduce

4y = 76

y = 19

We then plug into 4y + 34 to get angle

WZX = 110

5 0

2 years ago